How do we know that wave f. is the eigenfunction of an operator H?

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Main Question or Discussion Point

I am kind of new to this eigenvalue, eigenfunction and operator things, but i have come across this quote many times:
##\psi## is the eigenfunction of an operator ##\hat{H}## with eigenvalue ##W##.
First i need some explaination on how do we know this? All i know about operator ##\hat{H}## so far is this equation where ##\langle W \rangle## is an energy expected value:

\begin{align}
\langle W \rangle &= \int \limits_{-\infty}^{\infty} \overline{\Psi}\, \left(- \frac{\hbar^2}{2m} \frac{d^2}{d \, x^2} + W_p\right) \Psi \, d x
\end{align}

From which it follows that ##\hat{H} = - \frac{\hbar^2}{2m} \frac{d^2}{d \, x^2} + W_p##.

Aditional question:

I know how to derive relation ##\hat{H}\hat{a} = (W - \hbar \omega)\hat{a} \psi## for which they state that:

##\hat{a} \psi## is an eigenfunction of operator ##\hat{H}## with eigenvalue ##(W-\hbar \omega)##.
I also know how to derive relation ##\hat{H}\hat{a}^\dagger = (W + \hbar \omega)\hat{a}^\dagger \psi## for which they state that:

##\hat{a}^\dagger \psi## is an eigenfunction of operator ##\hat{H}## with eigenvalue ##(W+\hbar \omega)##.
How do we know this?
 

Answers and Replies

  • #2
HallsofIvy
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That's pretty much the definition of "eigenfunction" and "eigenvalue", isn't it?
 
  • #3
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That's pretty much the definition of "eigenfunction" and "eigenvalue", isn't it?
What exactly here is the definition? Which equation?
 
  • #4
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What exactly here is the definition? Which equation?
Any equation of the form
##\hat{H} \psi = E \psi##
is an eigen-equation

The functions ##\psi_n## which satisfy this equation are called the eigenfunctions of the operator ##\hat{H}##, with corresponding eigenvalues ##E_n##.

Put simply, when an operator ##\hat{H}## acts upon one of its eigenfunctions, it returns the same eigenfunction multiplied by a number, which we call the eigenvalue.
 
  • #5
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Any equation of the form
##\hat{H} \psi = E \psi##
is an eigen-equation

The functions ##\psi_n## which satisfy this equation are called the eigenfunctions of the operator ##\hat{H}##, with corresponding eigenvalues ##E_n##.

Put simply, when an operator ##\hat{H}## acts upon one of its eigenfunctions, it returns the same eigenfunction multiplied by a number, which we call the eigenvalue.
1st:
In this equation ##\hat{H}\hat{a} = (W - \hbar \omega)\hat{a} \psi## is it possible that ##\hat{a} \psi## is an eigenfunction... i mean ##\hat{a}## is an operator...

2nd:
And is it possible that my quotings were wrong i will write the wrong one and the one that i think is better:

WRONG:
##\hat{a}\psi## is an eigenfunction of operator ##\hat{H}## with eigenvalue ##(W−\hbar \omega)##.

RIGHT:
##\hat{a}\psi## is an eigenfunction of operators ##\hat{H}\hat{a}## with eigenvalue ##(W−\hbar \omega)##.
 
  • #6
Nugatory
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1st:
In this equation ##\hat{H}\hat{a} = (W - \hbar \omega)\hat{a} \psi## is it possible that ##\hat{a} \psi## is an eigenfunction... i mean ##\hat{a}## is an operator...

2nd:
And is it possible that my quotings were wrong i will write the wrong one and the one that i think is better:

WRONG:
##\hat{a}\psi## is an eigenfunction of operator ##\hat{H}## with eigenvalue ##(W−\hbar \omega)##.

RIGHT:
##\hat{a}\psi## is an eigenfunction of operators ##\hat{H}\hat{a}## with eigenvalue ##(W−\hbar \omega)##.
What's right is wrong and what's wrong is right :smile:
An operator maps one function into another, so if ##\hat{a}## is an operator and ##\psi## is a function, then ##\hat{a}\psi## is itself a function.

And I think that you dropped a ##\psi## from the equation you started with. I would have expected ##\hat{H}\hat{a}\psi = (W - \hbar \omega)\hat{a} \psi##, and written that way it is clear that the function ##\hat{a}\psi## is an eigenfunction of ##\hat{H}## with eigenvalue ##(W−\hbar \omega)##... Or at least adding a few parentheses will make it clear:

[tex]\hat{H}(\hat{a}\psi) = (W - \hbar \omega)(\hat{a} \psi)[/tex]
 
  • #7
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What's right is wrong and what's wrong is right :smile:
An operator maps one function into another, so if ##\hat{a}## is an operator and ##\psi## is a function, then ##\hat{a}\psi## is itself a function.

And I think that you dropped a ##\psi## from the equation you started with. I would have expected ##\hat{H}\hat{a}\psi = (W - \hbar \omega)\hat{a} \psi##, and written that way it is clear that the function ##\hat{a}\psi## is an eigenfunction of ##\hat{H}## with eigenvalue ##(W−\hbar \omega)##... Or at least adding a few parentheses will make it clear:

[tex]\hat{H}(\hat{a}\psi) = (W - \hbar \omega)(\hat{a} \psi)[/tex]
Ok so the eigenfunction can be a combination of an operator $\hat{a}$ and a function $\psi$?
 
  • #8
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Ok so the eigenfunction can be a combination of an operator $\hat{a}$ and a function $\psi$?
Why not? An operator acting on a function gives you another function.
 
  • #9
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Thanks.
 

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